Prove that f R rightarrow R given by fx x3 is continuous bu

Prove that f: R rightarrow R given by f(x) = x^3 is continuous but not uniformly continuous (it doesn\'t matter whether you use the definition of continuity given in the book, or whether you use yours truly\'s definition, which only applies to accumulation points). You can use all the material from chpt.0 to chpt.3.

Solution

Let\'s 1st proof that f(x) = x^3 is continuos for all R,

we know that for f(x) to be continuos it should satisfy |f(x) -f(x₀)| <= e for all x₀ belonging to R and a small postive integer e.

putting the values |x³x₀³|= |(x-x₀)(x²+x₀²+x*x₀)| ------------------------------------(1)

as you may know that |x||x₀|+|xx₀| ,

|x²+x*x₀+x₀²+|(+|x₀|)²+|x₀|+|x₀|²=²+3|x₀|+2|x₀|²

so |x³x₀³|(²+3|x₀|+2|x₀|²) and you can pick =min(1+3|x0|+2|x0|2,1) to prove its continuity.

*Note - Alternatively, you could just ignore the whole mess by noting that the product of continuous functions is continuous, so x³=xxx is continuous since x is (choose =).

Now I will prove that f(x) is not uniformly continuous by contradiction method

Suppose f(x)=x³ were uniformly continuous. Then for any , there is a that works for all a.

Let\'s take =1; we are now given such that for all a and for all x with |xa|<, we must have |f(x)f(a)|<.

Choose a large enough so that 3a²/2>1; for example a=((2/3)) +1 works.

Now take x=a+/2; this satisfies |xa|=/2<.

Hence we SHOULD have |f(x)f(a)|<=1.

Instead we have |f(x)f(a)|=|f(a+/2)f(a)|

                                     =|(a+/2)³a³|

                                     =|(3a²)/2+3a²/4+³/8|

                                      |(3a²)/2|>1.

This is a contradiction, and hence f(x)

is not uniformly continuous.